Pagination

syafid

Hi Guru.
I Really need yr help.
Im creating one page(page2.php) with select option.
After submit, this will go to other page (pagination.php) that display the selection result. But it have problem with next page.When I click next to page 2, it only display blank page not the rest of the selection result.
Here I the code :
pagination.php

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Pagination</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<form name="frm1" method="post" action="page2.php">
<?php
include('includes/open_db2.inc.php');

$sql = "SELECT * FROM prec_id ";
echo "Pilih Presint"."<br>";
$result = mysql_query($sql) or die(mysql_error());
$rows = mysql_num_rows($result);

//echo "<select name=\"id[]\">";
echo "<select name=id value=''></option>";
while ($rows=mysql_fetch_array($result)) {//mysql_fetch_array
  echo "<option value=\"{$rows["precint"]}\">{$rows["precint"]}</option>";
  //echo "<option value=$rows[ref]>$rows[plan_nm]</option>";

  }
echo "</select>";
echo "<br>";

?>
<input type="submit" name="submit" value="submit" />
</form>
</body>
</html>

and here the page2.php

<?php

include('includes/open_db2.inc.php');
echo $id2 = $_POST['id'];
//echo "<br>";
//echo $offset;
//$id2 = "presint 1";
$rows_per_page = 10;
//$page = 1;

//$limit = 'LIMIT ' .$rows_per_page;
/*calculate the offset */

//$limit = 'LIMIT ' .($pageno - 1) * $rows_per_page .',' .$rows_per_page;

if(isset($_GET['pageno'])) {
  $pageno = $_GET['pageno'];
  } else {
  $pageno = 1;

  }
echo $pageno; 
//echo $page;
$offset = ($pageno - 1) * $rows_per_page;


//if($_POST['submit'])
//{
$sql2 = "SELECT proj_nm FROM ho_title_proj where precint = '$id2' ORDER BY proj_nm LIMIT $offset, $rows_per_page";//where precint = '$id2' 
$result2 = mysql_query($sql2) or die(mysql_error());
/* */

/* */
while(list($proj_nm) = mysql_fetch_array($result2))
{
 echo "$proj_nm <br>";
}
echo '<br>';
//SELECT count(*) from ho_title_proj where precint = 'presint 1'
$query = "Select count(*) as proj_nm from ho_title_proj where precint = '$id2'";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result,MYSQL_ASSOC);
$numRows = $row['proj_nm'];

$maxpage = ceil($numRows/$rows_per_page);
$self = $_SERVER['PHP_SELF'];

if($pageno  < 1)
{
 $pageno = 1;
 }
 elseif($pageno > $maxpage)
  {
   $pageno = $maxpage;
   }

if($pageno > 1)
{
 $page = $pageno - 1 ;
 /*$prev = "<a href=\"$self?pageno=$page\">< Prev</a> ";
 $first = "<a href=\"$self?pageno=1\"><< First Page</a> "; //pageno=1
  original code*/
  $prev = "<a href='page2.php?pageno=$page'>< Prev</a> ";
  //"<a href='page2.php?pageno=$i'>$i</a>|";
  $first = "<a href='page2.php?pageno=1'><< First Page</a> "; //pageno=1
}
else
{
 $prev = '&nbsp;';//'< Prev';
 $first = '&nbsp;';//'<< First Page';
 }

if ($pageno < $maxpage)
{
 $page = $pageno + 1;//$pageno
 /*$next = " <a href=\"$self?pageno=$page\">Next ></a> ";//$page
 $last = " <a href=\"$self?pageno=$maxpage\">Last Page >></a> ";
  original code*/
  $next = " <a href='page2.php?pageno=$page'>Next ></a> ";//$page
  $last = " <a href='page2.php?pageno=$maxpage'>Last Page >></a> ";
 }
 else
 {
  $next = '&nbsp;';//'Next >';
  $last = '&nbsp;';//'Last Page >>';

  }

  echo $first . $prev . " Showing page $pageno of $maxpage pages " . $next . $last;

/*  for($i=1; $i <= $maxpage; $i++) {
   if($pageno == $i)
     echo $i."|";
   else
     echo "<a href='page2.php?pageno=$i'>$i</a>|";
	}*/

//  }
?>

Help me guru to rectify which line is problem..
Thanks

NogDog

Please use

 tags.[/url]

seisei

try to check the logs on your apache folder and see the error log textfile it will tell you what errors occur while you are running the apache server. it will also tells what specific line causes error.

syafid

seisei wrote:
try to check the logs on your apache folder and see the error log textfile it will tell you what errors occur while you are running the apache server. it will also tells what specific line causes error.

actually, it's no error. Just wondering why the output doesnt show.
if I'm not using pagination.php to post id to page2.php, and set manually
id='presint 2' at page2.php, the output is there. The output is show in 6 page by 10 rows per pages.

Can anybody advise me ?

leatherback

No, it is not an error. There is a reasoning error. You never send the ID of the sets that you want from the database. In the first page you post this through a <select>. However, when you then click a link to go to the next page, this variable it not defined anymore. Include it in your links, or place it in a session, and you can access it againon the next pages, and off you go.

seisei

try to place onchange='submit();' on your select tags at the pagination.php page and put it in session like $SESSION['index']=$POST['id'] (['id'] is the name ot the select tag)
and in your page2.php try to add
$index = $_SESSION['index'];

if (isset($index ){
$sql2 = "SELECT proj_nm FROM ho_title_proj where precint = '$id2' ORDER BY proj_nm LIMIT $offset, $rows_per_page";
}
else{
$sql2 = "SELECT proj_nm FROM ho_title_proj where precint = '$id2' ORDER BY proj_nm LIMIT 0, $rows_per_page";
}

leatherback

seisei wrote:
try to place onchange='submit();' on your select tags at the pagination.php page

Note that this will make the page fully dependent on javascript to function;
Also: It will always refresh a page for people with javascript, even if they are just checking the options in the select, and happen to select a different option.

seisei wrote:
$index = $_SESSION['index'];

if (isset($index ){
$sql2 = "SELECT proj_nm FROM ho_title_proj where precint = '$id2' ORDER BY proj_nm LIMIT $offset, $rows_per_page";
}
else{
$sql2 = "SELECT proj_nm FROM ho_title_proj where precint = '$id2' ORDER BY proj_nm LIMIT 0, $rows_per_page";
}

Instead of writing the query twice (Requiring a double update when you modify your code, only change the offset:

array_key_exists('index', $_SESSION) ? $offset = $_SESSION['index'] : 0;

$sql2 = "SELECT proj_nm FROM ho_title_proj where precint = '$id2' ORDER BY proj_nm LIMIT $offset, $rows_per_page";

seisei

<select size="1" class="text" name="pageindex" onChange="submit();">
i had this onChange="submit();" in my php project but i didnt use any javascript, its like a submit button in a form of select. when i click the value in my select fields it just submit it. hekhek and ofcourse i dont know bout the

"array_key_exists('index', $SESSION) ? $offset = $SESSION['index'] : 0;

cause i was new in php hehe i just self study 1 month ago and im answering question to the best knowledge that i have based on my previous php project that i manage to make work hehehe. i guess syafid you should better listen to leatherback cause his a pro. hehehehehe sorry!

leatherback

Hey SeiSei,

OnChange IS javascript. It just doesn't have the <> around it 😃

Ehm.. Pro.. Don't think so.. It is just a hobby 😃 But I have been at it a bit longer then you by the sound of it. I see new stuff every time I am on the forum, and sometimes I wonder why my code is so long and lumpy when I see solutions from others (e.g., look at laserlight, Bradgrafelman & weedpackets posts; sweet and short!). But that is how we learn, right? You try and you see someone else come up with an even better solution.

seisei

hekhek thank you for sharing it to me leatherback! now i have added info that i can use while coding. your right i learn from diffirent posts in the forum and super thanks to it!! hahaha!

syafid

leatherback wrote:
No, it is not an error. There is a reasoning error. You never send the ID of the sets that you want from the database. In the first page you post this through a <select>. However, when you then click a link to go to the next page, this variable it not defined anymore. Include it in your links, or place it in a session, and you can access it againon the next pages, and off you go.

Hi leatherback,

Thanks for the reply.
Yes yr right. I try to put the select value in the session but when i click next, i still get nothing.
Here I add some code at page2.php

session_start();
ob_start();
$id2 = $_POST['id'];
$_SESSION['name'] = $id2;

if(isset($_SESSION['name'])) {
$sql2 = "select proj_nm from ho_title_proj where precint = '$id2' order by proj_nm limit $offset, $rows_per_page";
} else {
$sql2 = "select proj_nm from ho_title_proj where precint = '$id2' order by proj_nm limit 0, $rows_per_page";
}
$result2 = mysql_query($sql2) or die(mysql_error());

Otherwise i would like to pass the precint id the url when click next but I have no idea..Can i pass 2 value from url to another page?

Please help and guide.

seisei

yes you can pass value by using &. example... page2.php?id=1&name=myname.
hehe i guess that is what your asking for. am i right leatherback?? hehe

syafid

seisei wrote:
yes you can pass value by using &. example... page2.php?id=1&name=myname.
hehe i guess that is what your asking for. am i right leatherback?? hehe

Hi SeiSei,
Thanks for reply. I used yr solution.
This is the code in page2.php

$id2 = $_SESSION['name'];
$next = "<a href='$self?pageno=$page&id=$id2'>Next ></a>";

At the top of the page2.php, im using $_GET['id'] to retrive the id for next page.
I combine all the solution that you and leatherback suggest and it's work.
FYI, I'm also a try n error programming. learn from expert people and guru.
Thanks all..

seisei

yes that ill always do. LEARN not just from the expert but from everybody hehe!